\(\int \frac {x^8}{a x+b x^3+c x^5} \, dx\) [78]

Optimal result

Integrand size = 20, antiderivative size = 100 \[ \int \frac {x^8}{a x+b x^3+c x^5} \, dx=-\frac {b x^2}{2 c^2}+\frac {x^4}{4 c}+\frac {b \left (b^2-3 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3} \]

[Out]

-1/2*b*x^2/c^2+1/4*x^4/c+1/4*(-a*c+b^2)*ln(c*x^4+b*x^2+a)/c^3+1/2*b*(-3*a*c+b^2)*arctanh((2*c*x^2+b)/(-4*a*c+b
^2)^(1/2))/c^3/(-4*a*c+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1599, 1128, 715, 648, 632, 212, 642} \[ \int \frac {x^8}{a x+b x^3+c x^5} \, dx=\frac {b \left (b^2-3 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}-\frac {b x^2}{2 c^2}+\frac {x^4}{4 c} \]

[In]

Int[x^8/(a*x + b*x^3 + c*x^5),x]

[Out]

-1/2*(b*x^2)/c^2 + x^4/(4*c) + (b*(b^2 - 3*a*c)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^3*Sqrt[b^2 - 4*
a*c]) + ((b^2 - a*c)*Log[a + b*x^2 + c*x^4])/(4*c^3)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 715

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2,
0] && NeQ[2*c*d - b*e, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 1128

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 1599

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^7}{a+b x^2+c x^4} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x^3}{a+b x+c x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {b}{c^2}+\frac {x}{c}+\frac {a b+\left (b^2-a c\right ) x}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b x^2}{2 c^2}+\frac {x^4}{4 c}+\frac {\text {Subst}\left (\int \frac {a b+\left (b^2-a c\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c^2} \\ & = -\frac {b x^2}{2 c^2}+\frac {x^4}{4 c}-\frac {\left (b \left (b^2-3 a c\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}+\frac {\left (b^2-a c\right ) \text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3} \\ & = -\frac {b x^2}{2 c^2}+\frac {x^4}{4 c}+\frac {\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}+\frac {\left (b \left (b^2-3 a c\right )\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^3} \\ & = -\frac {b x^2}{2 c^2}+\frac {x^4}{4 c}+\frac {b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.93 \[ \int \frac {x^8}{a x+b x^3+c x^5} \, dx=\frac {c x^2 \left (-2 b+c x^2\right )-\frac {2 b \left (b^2-3 a c\right ) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3} \]

[In]

Integrate[x^8/(a*x + b*x^3 + c*x^5),x]

[Out]

(c*x^2*(-2*b + c*x^2) - (2*b*(b^2 - 3*a*c)*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (b^2
- a*c)*Log[a + b*x^2 + c*x^4])/(4*c^3)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05

method	result	size
default	\(-\frac {-\frac {1}{2} c \,x^{4}+b \,x^{2}}{2 c^{2}}+\frac {\frac {\left (-a c +b^{2}\right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (a b -\frac {\left (-a c +b^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 c^{2}}\)	\(105\)
risch	\(\frac {x^{4}}{4 c}-\frac {b \,x^{2}}{2 c^{2}}+\frac {b^{2}}{4 c^{3}}-\frac {\ln \left (\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}+\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, a \right ) a^{2}}{c \left (4 a c -b^{2}\right )}+\frac {5 \ln \left (\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}+\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, a \right ) a \,b^{2}}{4 c^{2} \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}+\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, a \right ) b^{4}}{4 c^{3} \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}+\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, a \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}}{4 c^{3} \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}-\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, a \right ) a^{2}}{c \left (4 a c -b^{2}\right )}+\frac {5 \ln \left (\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}-\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, a \right ) a \,b^{2}}{4 c^{2} \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}-\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, a \right ) b^{4}}{4 c^{3} \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}-\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, a \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}}{4 c^{3} \left (4 a c -b^{2}\right )}\)	\(957\)

[In]

int(x^8/(c*x^5+b*x^3+a*x),x,method=_RETURNVERBOSE)

[Out]

-1/2/c^2*(-1/2*c*x^4+b*x^2)+1/2/c^2*(1/2*(-a*c+b^2)/c*ln(c*x^4+b*x^2+a)+2*(a*b-1/2*(-a*c+b^2)*b/c)/(4*a*c-b^2)
^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 313, normalized size of antiderivative = 3.13 \[ \int \frac {x^8}{a x+b x^3+c x^5} \, dx=\left [\frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} - 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x^{2} - {\left (b^{3} - 3 \, a b c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} - 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x^{2} + 2 \, {\left (b^{3} - 3 \, a b c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \]

[In]

integrate(x^8/(c*x^5+b*x^3+a*x),x, algorithm="fricas")

[Out]

[1/4*((b^2*c^2 - 4*a*c^3)*x^4 - 2*(b^3*c - 4*a*b*c^2)*x^2 - (b^3 - 3*a*b*c)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 +
2*b*c*x^2 + b^2 - 2*a*c - (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) + (b^4 - 5*a*b^2*c + 4*a^2*c^
2)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4), 1/4*((b^2*c^2 - 4*a*c^3)*x^4 - 2*(b^3*c - 4*a*b*c^2)*x^2 + 2*(
b^3 - 3*a*b*c)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + (b^4 - 5*a*b^2*c +
4*a^2*c^2)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (92) = 184\).

Time = 1.75 (sec) , antiderivative size = 391, normalized size of antiderivative = 3.91 \[ \int \frac {x^8}{a x+b x^3+c x^5} \, dx=- \frac {b x^{2}}{2 c^{2}} + \left (- \frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{4 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right ) \log {\left (x^{2} + \frac {2 a^{2} c - a b^{2} + 8 a c^{3} \left (- \frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{4 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right ) - 2 b^{2} c^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{4 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{4 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right ) \log {\left (x^{2} + \frac {2 a^{2} c - a b^{2} + 8 a c^{3} \left (\frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{4 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right ) - 2 b^{2} c^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{4 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \frac {x^{4}}{4 c} \]

[In]

integrate(x**8/(c*x**5+b*x**3+a*x),x)

[Out]

-b*x**2/(2*c**2) + (-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4*c**3))*log
(x**2 + (2*a**2*c - a*b**2 + 8*a*c**3*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c -
b**2)/(4*c**3)) - 2*b**2*c**2*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4
*c**3)))/(3*a*b*c - b**3)) + (b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4*c
**3))*log(x**2 + (2*a**2*c - a*b**2 + 8*a*c**3*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) -
(a*c - b**2)/(4*c**3)) - 2*b**2*c**2*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b
**2)/(4*c**3)))/(3*a*b*c - b**3)) + x**4/(4*c)

Maxima [F]

\[ \int \frac {x^8}{a x+b x^3+c x^5} \, dx=\int { \frac {x^{8}}{c x^{5} + b x^{3} + a x} \,d x } \]

[In]

integrate(x^8/(c*x^5+b*x^3+a*x),x, algorithm="maxima")

[Out]

1/4*(c*x^4 - 2*b*x^2)/c^2 - integrate(-((b^2 - a*c)*x^3 + a*b*x)/(c*x^4 + b*x^2 + a), x)/c^2

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.92 \[ \int \frac {x^8}{a x+b x^3+c x^5} \, dx=\frac {c x^{4} - 2 \, b x^{2}}{4 \, c^{2}} + \frac {{\left (b^{2} - a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} - \frac {{\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{3}} \]

[In]

integrate(x^8/(c*x^5+b*x^3+a*x),x, algorithm="giac")

[Out]

1/4*(c*x^4 - 2*b*x^2)/c^2 + 1/4*(b^2 - a*c)*log(c*x^4 + b*x^2 + a)/c^3 - 1/2*(b^3 - 3*a*b*c)*arctan((2*c*x^2 +
b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^3)

Mupad [B] (verification not implemented)

Time = 8.58 (sec) , antiderivative size = 842, normalized size of antiderivative = 8.42 \[ \int \frac {x^8}{a x+b x^3+c x^5} \, dx=\frac {x^4}{4\,c}-\frac {\ln \left (c\,x^4+b\,x^2+a\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}-\frac {b\,x^2}{2\,c^2}+\frac {b\,\mathrm {atan}\left (\frac {2\,c^4\,\left (4\,a\,c-b^2\right )\,\left (\frac {\frac {b\,\left (3\,a\,c-b^2\right )\,\left (\frac {8\,a^2\,c^4-8\,a\,b^2\,c^3}{c^4}-\frac {8\,a\,c^2\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )}{8\,c^3\,\sqrt {4\,a\,c-b^2}}-\frac {a\,b\,\left (3\,a\,c-b^2\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{c\,\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}}{a}-x^2\,\left (\frac {\frac {b\,\left (\frac {6\,b^3\,c^3-10\,a\,b\,c^4}{c^4}+\frac {4\,b\,c^2\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (3\,a\,c-b^2\right )}{8\,c^3\,\sqrt {4\,a\,c-b^2}}+\frac {b^2\,\left (3\,a\,c-b^2\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{2\,c\,\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}}{a}+\frac {b\,\left (\frac {2\,a^2\,b\,c^2-3\,a\,b^3\,c+b^5}{c^4}+\frac {\left (\frac {6\,b^3\,c^3-10\,a\,b\,c^4}{c^4}+\frac {4\,b\,c^2\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}-\frac {b^3\,{\left (3\,a\,c-b^2\right )}^2}{2\,c^4\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )+\frac {b\,\left (\frac {\left (\frac {8\,a^2\,c^4-8\,a\,b^2\,c^3}{c^4}-\frac {8\,a\,c^2\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}-\frac {a^3\,c^2-2\,a^2\,b^2\,c+a\,b^4}{c^4}+\frac {a\,b^2\,{\left (3\,a\,c-b^2\right )}^2}{c^4\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )}{9\,a^2\,b^2\,c^2-6\,a\,b^4\,c+b^6}\right )\,\left (3\,a\,c-b^2\right )}{2\,c^3\,\sqrt {4\,a\,c-b^2}} \]

[In]

int(x^8/(a*x + b*x^3 + c*x^5),x)

[Out]

x^4/(4*c) - (log(a + b*x^2 + c*x^4)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a*c^4 - 4*b^2*c^3)) - (b*x^2)/(2*
c^2) + (b*atan((2*c^4*(4*a*c - b^2)*(((b*(3*a*c - b^2)*((8*a^2*c^4 - 8*a*b^2*c^3)/c^4 - (8*a*c^2*(2*b^4 + 8*a^
2*c^2 - 10*a*b^2*c))/(16*a*c^4 - 4*b^2*c^3)))/(8*c^3*(4*a*c - b^2)^(1/2)) - (a*b*(3*a*c - b^2)*(2*b^4 + 8*a^2*
c^2 - 10*a*b^2*c))/(c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a - x^2*(((b*((6*b^3*c^3 - 10*a*b*c^4)/c^4
+ (4*b*c^2*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(3*a*c - b^2))/(8*c^3*(4*a*c - b^2)^(1/2)
) + (b^2*(3*a*c - b^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a +
(b*((b^5 + 2*a^2*b*c^2 - 3*a*b^3*c)/c^4 + (((6*b^3*c^3 - 10*a*b*c^4)/c^4 + (4*b*c^2*(2*b^4 + 8*a^2*c^2 - 10*a
*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a*c^4 - 4*b^2*c^3)) - (b^3*(3*a*c -
b^2)^2)/(2*c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))) + (b*((((8*a^2*c^4 - 8*a*b^2*c^3)/c^4 - (8*a*c^2*(2
*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a*c^4 - 4*b^2
*c^3)) - (a*b^4 + a^3*c^2 - 2*a^2*b^2*c)/c^4 + (a*b^2*(3*a*c - b^2)^2)/(c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2
)^(1/2))))/(b^6 + 9*a^2*b^2*c^2 - 6*a*b^4*c))*(3*a*c - b^2))/(2*c^3*(4*a*c - b^2)^(1/2))